Problem: Define $f(x, y) = x^2y$. Let $\vec{a} = (3, -1)$ and $\vec{v} = \left( 1, -3 \right)$. Calculate $ \lim_{h \to 0} \dfrac{f(\vec{a} + h \vec{v}) - f(\vec{a})}{h}$.
Explanation: Our ultimate goal is to substitute $h = 0$ into the limit and get the answer. To do this, we first need to cancel out the $h$ in the denominator. Otherwise we're dividing by zero. How can we simplify the limit so that we get an $h$ on the top and an $h$ on the bottom, which we can then cancel? Let's plug in $\vec{a}$ and $\vec{v}$ to the limit. $ \lim_{h \to 0} \dfrac{f \left( (3, -1) + h \left( 1, -3 \right) \right) - f(3, -1)}{h}$ Now we can add together the vectors. $ \lim_{h \to 0} \dfrac{f \left( 3 + h, -1 - 3h \right) - f(3, -1)}{h}$ Let's evaluate $f$. $ \lim_{h \to 0} \dfrac{(3 + h)^2(-1 - 3h) - 9(-1)}{h}$ When we simplify, we can cancel out all the terms without an $h$. $\begin{aligned} & \lim_{h \to 0} \dfrac{(3 + h)^2(-1 - 3h) - 9(-1)}{h} \\ \\ &= \lim_{h \to 0} \dfrac{(9 + 6h + h^2)(-1 - 3h) + 9}{h} \\ \\ &= \lim_{h \to 0} \dfrac{-9 - 6h - h^2 - 27h + 18h^2 -3h^3 + 9}{h} \\ \\ &= \lim_{h \to 0} \dfrac{- 33h + 17h^2 -3h^3}{h} \end{aligned}$ Now we can cancel $h$ and calculate the limit. $\begin{aligned} \lim_{h \to 0} \dfrac{-33h + 17h^2 - 3h^3}{h} &= \lim_{h \to 0} -33 + 17h - 3h^2 \\ \\ &= -33 \end{aligned}$ In conclusion, $ \lim_{h \to 0} \dfrac{f(\vec{a} + h \vec{v}) - f(\vec{a})}{h} = -33$.